Pythagoras is giving yet another lecture that All is Number... (Here, a number is a rational number p/q)
All is Number.
...while using invalid proof methods yet again.
Why? Consider any quantity. We know we can represent the whole part as a number trivially. For the fractional component, we can approximate the quantity with arbitrary precision as follows: call the desired quantity x (x 1). For each integer i 0, if x = 2^(-i), set x to x - 2^(-i) and add 1/2^i, a number. We know that a number plus a number is a number so we have only used numbers so far. After a given i, we know x 2^(-i), so any quantity can be approximated to an arbitrary precision. Therefore, it is absurd to think a quantity exists that is not a number. Not only that, but numbers are also intrinsic in nature. Consider musical notes produced from a vibrating string. A string 1/2 as long produces an octave, 2/3 produces a fifth, and 3/4 produces a fourth. Thus, numbers are a fundamental part of nature. If not all is number, what is it...
Hippasus pursued the truth. He toiled for many days, eventually disproving the statement.
All is not number! I have discovered a truly marvelous proof of this which this margin is too narrow to contain.
So I'll use another book I guess...
Hippasus slowly realized the other implications, including the painful truth that what he believed his whole life was false.
I have discovered that sqrt(2) is not a number. Call quantities that make use of the terms x^(1/y), where y is a natural number, algebraic numbers. Are those all quantities that exist? What if there exist quantities that transcend even that? Transcendentals?
More importantly, this means everything I know is false. Has my life until now really been for anything? And the lives of my fellow students as well...
However, now I can see the ray of sunlight at the end of the cave! We can correct our doctrine. All we have to do is place our absolute trust in proofs!
Hippasus returned and presented his revelation.
I will prove sqrt(2) is not a number by contradiction. Let sqrt(2) = p/q, where p and q are coprime. qsqrt(2) = p. 2q^2 = p^2. This implies p is even, so let 2k = p. 2q^2 = (2k)^2. 2q^2 = 4k^2. q^2 = 2k^2. This implies q is even. But since p is also even, p and q share a factor, namely 2, a contradiction. QED.
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